Have you tried exploring much smaller powers of each of those numbers?

111^1 = 111
111^2 = 12321
111^3 = 1367631
...
111^6 = 1870414552161

Seems relatively safe to say that 111^n (where n is an integer) ends in a 1. Makes sense, considering the ones digit is always going to be a product of previous 1s digits, which are all 1. No matter how many 1s you multiply together, you still end up with a 1s digit of 1.

Now, 333 is a little trickier, since multiplying a bunch of 3s together gets you a number that ends in: